1. Two Way Slab
Problem Statement:
Use the ACI 318 Direct Design Method to design an interior bay of a flat plate slab system of
multi bay building. The Dimensions of an interior bay are shown in Figure 1.
20'
24'
Figure 1. Interior Bay Dimensions
This example is to be used for educational purposes only. 1

2. Design Data
Column Size - 24” x 24”
Materials
• Concrete: normal weight (150 pcf), f’c = 4,000 psi
• Welded Wire Reinforcement (WWR) , fy = 80,000 psi
Loads
• Superimposed dead loads = 30 psf
• Live load = 50 psf
Slab Thickness
Longest clear span ln = 24 – (24/12) = 22.0 ft
Minimum thickness h per ACI Table 9.5(c) = ln/30 = 8.8 in. Use h 9.0 in.
qu = 1.2(9/12*150 + 30) + 1.6(50) = 251.0 psf
Design for Flexure
Check if the Direct Design Method of ACI Sect. 13.6 can be utilized to compute the bending
moments due to the gravity loads:
• 3 continuous spans in one direction, more than 3 in the other O.K.
• Rectangular panels with long-to-short span ratio = 24/20 = 1.2 < 2 O.K.
• Successive span lengths in each direction are equal O.K.
• No offset columns O.K.
• L/D = 50/(112.5 + 30) = 0.35 < 2 O.K.
Since all requirements are satisfied, the Direct Design Method can be used.
Short Direction (20 ft)
l2 = 24 ft and ln = 20’-2’= 18.0 ft
Factored moment per span.
Mo = =
Division of the total panel moment Mo into negative and positive moments, and then column and
middle strip moments, involves the direct application of the moment coefficients in the following
Table 1.
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3. Slab Moment Int. Neg. Int. pos.
Total Moment 170.80 85.40
Column Strip 128.10 51.24
Middle Strip 42.70 34.16
Table 1 Moment Distribution.
Reinforcement,
Assume D16 wires will be used.
d = 9 - 0.75 -0.45/2 = 8.03 in.
Column strip width b = (20 x 12)/2 = 120 in.
Middle strip width b = (24 x 12) – 120 = 168 in.
As = Mu /4d*60/80 where Mu is in ft-kips and d is in inches
Min. As = 0.0018bh*60/80 (grade 80,000 psi steel)
Max. s = 2h = 18 in. or 18 in. (Sect. 13.3.2)
The required reinforcements are shown in table 2.
Span Location Mu kips-ft b (in) d (in) As (in2) Min As (in2)
Reinforcement
Column Int. Neg. 128.10 120.00 8.03 2.99 1.46 D16 at 6 in.
Strip Int. Pos. 51.24 120.00 8.03 1.20 1.46 D8 at 6 in.
Middle Int. Neg. 42.70 168.00 8.03 1.00 2.04 D8 at 6 in.
Strip Int. Pos. 34.16 168.00 8.03 0.80 2.04 D8 at 6 in.
Table 2. Required Bay Reinforcement.
Long Direction (24 ft)
l2 = 20 ft and ln = 24-2= 22.0 ft
Factored moment per span.
Mo = =
Division of the total panel moment Mo into negative and positive moments, and then column and
middle strip moments, involves the direct application of the moment coefficients in the following
Table 3.
Slab Moment Int. Neg Int. Pos.
Total moment 212.10 106.05
Column Strip 159.08 63.63
Middle Strip 53.03 42.42
Table 3 Moment Distribution.
This example is to be used for educational purposes only. 3

4. Reinforcement,
Assume D20 wires will be used.
d = 9 - 0.75 -0.45- 0.5/2 = 7.55 in.
Column strip width b = (20 x 12)/2 = 120 in.
Middle strip width b = (20 x 12) – 120 = 120 in.
As = Mu /4d*60/80 where Mu is in ft-kips and d is in inches
Min. As = 0.0018bh*60/80 (grade 80 ksi )
Max. s = 2h = 18 in. or 18 in. (Sect. 13.3.2)
The required reinforcements are shown in table 4.
Span Location Mu kips-ft b (in) d (in) As (in2) Min As (in2)
Reinforcement
Column Int. Neg. 159.08 120.00 7.55 3.95 1.46 D20 at 6 in.
Strip Int. Pos. 63.63 120.00 7.55 1.58 1.46 D8 at 6 in.
Middle Int. Neg. 53.03 120.00 7.55 1.32 1.46 D8 at 6 in.
Strip Int. Pos. 42.42 120.00 7.55 1.05 1.46 D8 at 6 in.
Table 4. Required Bay Reinforcement.
Reinforcement Details:
The following figures 2, 3, 4 and 5show the reinforcement details for the column and middle
strips for both directions. The lengths were determined from Figure 13.3.8 of ACI 318-05.
WWR2
WWR3 WWR1
Top Reinforcement
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5. 20'
24'
WWR4
10'-8"
10'-8" 10'-8"
Bottom Reinforcement
Figure 2. Top and Bottom Reinforcement
11'
9'
13' 11'-8"
D8@6"
D20@12"
D16@12" 2 D8
WWR1 WWR2
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6. 10'-8"
10'
11' 2- D8 26'
D8@6"
WWR3 D8@6"
WWR4 D8@6"
Figure 3. Sheets Details
2'
11' 3
5'-44" 3
5'-44"
11' WWR1 WWR1
10'-8"
10'-8" 10'-8"
20'
WWR4
Column Strip
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7. 10' 10'
WWR3
10'-8"
10'-8" 10'-8"
20'
WWR4
Middle Strip
Figure 4. Cross Section (Short Direction)
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8. 13' 6'-71" 6'-71"
13' 4 2' 4
WWR1 WWR1
WWR4
24'
26'
Column Strip
11'-8" 11'-8"
WWR2
WWR4
24'
26'
Middle Strip
Figure 5. Cross Section (Long Direction)
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9. Reinforcement Quantities:
Weight Total
per Foot Weight of
Sheet Number and Size Length (ft) (lb) Weight (lb) Sheet
11 D16 11 0.544 65.8
WWR1 14 D20 13 0.68 123.8 189.6
4 D8 9 0.272 9.8
WWR2 18 D8 11.67 0.272 57.1 66.9
22 D8 10 0.272 59.8
WWR3 4 D8 11 0.272 12.0 71.8
52 D8 10.67 0.272 150.9
WWR4 21 D8 26 0.272 148.5 299.4
Each 20’ by 24’ bay has one WWR1, one WWR2, one WWR3 and two WWR4. The total steel
weight for each bay = 189.6+66.9+71.8+2*299.4 = 927.1 lbs.
Steel weight per square foot = 927.1 / (24*20) = 1.93 lb/ft2
Design Using Grade 60 Rebar.
Figures 6 and 7 show the reinforcement detail of the interior bay using Grade 60 Rebar.
The total steel weight for each bay using Grade 60 Rebar = 1,204 lbs.
Steel weight per square foot = 1,204/(24*20) = 2.51 lb/ft2
Conclusions:
The ratio between the total weight per square foot using WWR to the total weight per square foot
using Grade 60 Rebar = (1.93/2.51) *100 = 77%.
Placed of WWR Should be significantly faster than placement of the reinforcing bars, resulting
in additional overall saving.
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10. 5
12'-98" 5
12'-98"
3
9'-28" 3
9'-28"
10 # 4 10 #4
10# 4
20'
Column Strip
10' 10'
14# 4
6"
3 7#4 7#4
2'-88"
20'
Middle Strip
Figure 6. Cross Section (Short Direction), Grade 60 Rebar
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11. 3
15'-28" 3
15'-28"
5
10'-98" 5"
10'-98
14 # 4 14 # 4
11# 4
24'
26'
Column Strip
11'-8" 11'-8"
10#4
6"
5
3'-38" 5#4 5#4
24'
Middle Strip
Figure 7. Cross Section (Long Direction), Grade 60 Rebar
This example is to be used for educational purposes only. 11